${\displaystyle \underset{1}{\overset{16}{\int }}{\tan }^{-1}\sqrt{\sqrt{x}-1}dx=}$

$\sqrt{x}-1=t^2,x={\left(t^2+1\right)}^2$ 

$I=\underset{0}{\overset{\sqrt{3}}{\int }}{\tan }^{-1}t d{(t^2+1)}^2$ $={\tan }^{-1}t{\cdot {\left(t^2+1\right)}^2|}_0^{\sqrt{3}}-{\displaystyle \underset{0}{\overset{\sqrt{3}}{\int }}\left(t^2+1\right)dt}$        by parts

                                            $=\frac{16\pi }{3}-{\left(\frac{t^3}{3}+t\right)}_0^{\sqrt{3}}=\frac{16\pi }{3}-2\sqrt{3}$